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3 years ago

title="( {8x}^{2} + 4x + 4) - (7x + 9)" alt="( {8x}^{2} + 4x + 4) - (7x + 9)" align="absmiddle" class="latex-formula">
Solve.

Mathematics

1 answer:

Rama09 [41]3 years ago

8 0

Answer:

Combining like terms together we get 4x - 7x = -3x and 4 - 9 = -5 so the answer is 8x² - 3x - 5.

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Automobiles arrive at a vehicle equipment inspection station according to a Poisson process with rate α = 8 per hour. Suppose th

Elis [28]

Answer:

a) The probability that exactly eight arrive during the hour and all eight have no violations is 0.0005.

b) For any fixed y ≥ 8, the probability that y arrive during the hour, of which eight have no violations is:

$P(X=y\,\&\,nv=8)=\frac{4^ye^{-8}}{8!(y-8)!}$

c) The probability that eight "no-violation" cars arrive during the next hour is 0.030.

Step-by-step explanation:

a) The probability that exactly eight arrive during the hour and all eight have no violations is equal to the product of the probability of arrival of 8 vehicules and the probability of having 8 vehicules with no violations.

$P(X=8\,\&\,no\, violations)=P(no\, violations|X=8)*P(X=8)\\\\P(X=8\,\&\,nv)=(0.5)^8*\frac{8^8e^{-8}}{8!} = 0.0039*0.1396=0.0005$

b) For any fixed y ≥ 8, the probability that y arrive during the hour, of which eight have no violations is:

$P(X=y\,\&\,nv=8)=P(nv=8|X=y)*P(X=y)\\\\P(X=y\,\&\,nv=8)=[\binom{y}{8}(0.5)^8*(0.5)^{y-8}]*\frac{8^ye^{-8}}{y!} =\frac{y!}{8!(y-8)!}0.5^y *8^y*\frac{e^{-8}}{y!}\\\\ P(X=y\,\&\,nv=8)=(\frac{y!}{y!})(0.5*8)^y\frac{e^{-8}}{8!(y-8)!}=\frac{4^ye^{-8}}{8!(y-8)!}$

c) Using the result of point (b) we can express the probability that eight "no violation" vehicules arrive durting the next hour as:

$P(nv=8)=\sum\limits^\infty_{y=8} {\frac{4^ye^{-8}}{8!(y-8)!}}=\frac{e^{-8}}{8!} \sum\limits^\infty_{y=8} {\frac{4^y}{(y-8)!}}=\frac{e^{-8}4^{8}}{8!} \sum\limits^\infty_{y=8} {\frac{4^{y-8}}{(y-8)!}}\\\\P(nv=8)=\frac{e^{-8}4^{8}}{8!} \sum\limits^\infty_{z=0} {\frac{4^{z}}{z!}}=\frac{e^{-8}4^{8}}{8!}*e^4=\frac{e^{-4}4^{8}}{8!}\\\\P(nv=8)= 0.030$